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Finding the pattern

  1. Sep 12, 2005 #1
    Is there anyone who can help me? I need to find a pattern in these numbers:
    1/2, 1/6, 1/12, 1/20, 1/30, 1/42, 1/56. Now, I know just by looking at the denominators, if I could only work with those, I could use the formula:
    2n+a_n_1. But I have that fraction, so it's all screwy. Anyone see where I'm screwing up? I know I can't.
     
  2. jcsd
  3. Sep 12, 2005 #2

    TD

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    How about [tex]u_n = \frac{1}
    {{n\left( {n + 1} \right)}}[/tex]
     
  4. Sep 12, 2005 #3
    well, this is a summation problem. it is the summation of 1/n(n+1) equals "what." the "what" is what i'm supposed to find. the 1/n(n+1) is already given for the one side of the equation, i need to find what the summation is also equal to.
     
  5. Sep 12, 2005 #4
    have you try just doing a few terms by hand, it might give you a clue!!!!
     
  6. Sep 12, 2005 #5

    TD

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    I thought you were looking for a pattern to find the formula.
    Are you looking for a formula for a partial sum or for the sum of the infinite series?
     
  7. Sep 12, 2005 #6
    Do you honestly think I have not tried that. Give me some credit. :devil:
     
  8. Sep 12, 2005 #7

    TD

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    Have you thought of the expansion?

    [tex]\frac{1}
    {{n\left( {n + 1} \right)}} = \frac{1}
    {n} - \frac{1}
    {{n + 1}}[/tex]
     
  9. Sep 12, 2005 #8
    1/n(n+1)=1/n - 1/(n+1)

    If you write the sum: Sn= (1/2-1/6) + (1/6-1/12) + ... + (1/n - 1/(n+1) )
    All the terms cancel except 1/2 and 1/n+1
    i.e. Sn= 1/2 - 1/(n+1)= (n-1)/2(n+1)

    Taking the limit n->infinity, we get: 1/2

    I hope it's the correct answer!
     
  10. Sep 12, 2005 #9

    TD

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    The first element (n = 1) already is 1/2 so it has to be more. Other than that, your work looks good so it should be 1/2 + your 1/2 = 1 :smile:

    The partial sum is

    [tex]s_n = \frac{n}
    {{n + 1}}[/tex]

    So for the infinite series

    [tex]\mathop {\lim }\limits_{n \to \infty } s_n = \mathop {\lim }\limits_{n \to \infty } \frac{n}
    {{n + 1}} = 1[/tex]
     
  11. Sep 12, 2005 #10
    You are right. I made a mistake. The first term of Sn should've been '1', since 1/n gives 1 for n=1.

    So the series converges to 1, and all terms from 1/2 to 1/n cancel in the partial sum to give, Sn=n/(n+1), just as you point out.
     
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